Cayley graph on $ D_{2n} $ and $ \mathbb Z_n$

Question

How we can make Cayley graph on $ D_{2n} $ and $ \mathbb Z_n$? What can be S in $Cay(D_{2n},S)$ and $ Cay(\mathbb Z_n ,S)$, Please write one example. Thanks for advise.

Answer 1

Here's an example of a Cayley graph for the group $\mathbb{Z}_{10}$:

The vertices are the underlying set of the group $\mathbb{Z}_{10}$, and we draw edges between $g$ and $g+h$ whenever $g,h \in S$ where $S=\{\pm 1, \pm 2\}$.

Here's an example of a Cayley graph for the dihedral group $D_{10}$:

The vertices are the underlying set of the dihedral group with presentation $$D=\langle f,r | f^2=e, r^5=e, rf=fr^{-1} \rangle,$$ (we can think of $f$ as "flip" and $r$ as "rotation") and we draw edges between $g$ and $g+h$ whenever $g,h \in S$ where $S=\{f,r,r^{-1}\}$.

Answer 2

Recalling the definition of Cayley graph of a group $G$, remember that the vertex set is the set $G$ and two vertices, $g, h\in G$, are connected by an edge if there exists some $a\in S$ such that $g^{-1}h\in S$. So this means that $S\subset G$. Now in principal you could let $S$ be any subset of $G$ but usually (always?) you want to choose $S$ to be a generating set for $G$ because this ensures that the resulting graph is connected. So in paticular, since groups have many different generating sets, the resulting graphs will be different. So there is no such thing as $\textbf{the}$ Cayley graph of a group.

Just to give one example. If $G=\{e, g_1, ..., g_n\}$, note $|G|=n+1$, and if we let $S=G\setminus\{e\}$. Then Cayley($G, S)$= $K_{n+1}$, the complete graph on $n+1$ vertices.

However, in general you do not do this. Instead, you try to use the smallest $S$ possible. In particular for the case $\mathbb{Z}_n$ you can choose $S=\{1\}$. In this case you will get that the Cayley graph is a loop with $n$ points on it.

Answer 3

Considering the definition of Cayley graph done by @Owen, you can do it as follows:

For $\mathbb Z_n=\langle a\rangle=\{1,a,a^2,\cdots a^{n-1}\}$:

And for $D_{2n}$, I did it for $D_6=\{1,a,a^2,b,ab,a^2b\}$ of order $6$ instead: